Question

Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hole densities of 104/cm3 and 1016/cm3, respectively, when a 1 v is applied end-to-end. use μn = 1200 cm2/v · s and μp = 500 cm2/v · s.

Answer 1

Explanation :

It is given that,

Length of silicon bar, $l=10\mu m= 0.001\ cm$

Free electron density, $n_e=104\ cm^3$

Hole density, $n_h=1016\ cm^3$

$\mu_n=1200\ cm^2/Vs$

$\mu_p=500\ cm^2/Vs$

The total current flowing in bar is the sum of drift current due to hole and the electrons.

$J=J_e+J_h$

$J=nqE\mu_n+pqE\mu_p$

where, n and p are electron and hole densities.

$J=Eq[n\mu_n+p\mu_p]$

we know that $E=\dfrac{V}{l}$

So, $J=\dfrac{V}{l}q[n\mu_n+p\mu_p]$

$J=\dfrac{1.6\times 10^{-19}\ C}{0.001\ V}[104\ cm^{-3}\times 1200\ cm^2/V\ s+1016\ cm^{-3}\times 500\ cm^2/V\ s]$

$J=1012480\times 10^{-16}\ A/m^2$

or

$J=1.01\times 10^{-9}\ A/m^2$

Current, $I=JA$

A is the area of bar, $A= 20\mu m=0.002\ cm$

So, $I=1.01\times 10^{-9}\ A/m^2\times 0.002\ m^2$

Current, $I=2.02\times 10^{-12}\ A$

So, the current flowing in silicon bar is $I=2.02\times 10^{-12}\ A$.